Question

我想每隔一个元素向现有列表中插入一个元素。

让我们说列表是fruits = ["banana", "apple", "mango", "kiwi"]，添加元素"peach"就是我的事情：

fruits = ["banana", "apple", "mango", "kiwi"]
fruits_2 = list(fruits)

for i in range(len(fruits)):
    fruits_2.insert(2*i + 1, "peach")

print(fruits_2)

输出为

['banana', 'peach', 'apple', 'peach', 'mango', 'peach', 'kiwi', 'peach']

这是我想要的，但是我觉得可能有一种更好，更简洁的方法来执行此操作，而无需创建第二个列表。

Answer 1

使用zip以及itertools.repeat和itertools.chain。

>>> from itertools import chain, repeat
>>> fruits = ["banana", "apple", "mango", "kiwi"]
>>> list(chain.from_iterable(zip(fruits, repeat("peach"))))
['banana', 'peach', 'apple', 'peach', 'mango', 'peach', 'kiwi', 'peach']

repeat创建'peach'的无限序列。 zip创建一系列对，由水果中的一个项目和一个'peach'的实例组成。 chain.from_iterable将对的序列“平化”为单个序列，list从该序列中生成具体列表。

看中间步骤：

>>> from itertools import islice
>>> list(islice(repeat("peaches"), 5))
['peaches', 'peaches', 'peaches', 'peaches', 'peaches']

>>> zip(fruits, repeat("peaches"))
[('banana', 'peaches'), ('apple', 'peaches'), ('mango', 'peaches'), ('kiwi', 'peaches')]

这种方法可以归结为一系列appends到新列表，例如

result = []
for x in fruits:
    result.append(x)
    result.append("peaches")

但是，您无需从一对迭代器中获取参数，而不必将参数硬编码为append：

def peach_source():
    while True:
        yield "peach"

def fruit_source(fruits):
    for x in fruits:
        yield x

result = []
peaches = peach_source()
fruits = fruit_source(fruits)  # Yes, this is highly redundant; just showing the similarity to peach_source
done = False
while not done:
    try:
        result.append(next(fruits))
    except StopIteration:
        done = True
    result.append(next(peaches))

itertools.repeat为您创建peach_source。 zip处理获取水果和获取桃子之间的交替。 chain.from_iterable定义了附加到result的动作，而list实际上是执行附加。

Answer 2

仅使用“ 双倍长度”（以及偶数/奇数位置）魔术：

FROM (SELECT col_1, col_2, col_3 FROM t1)
INSERT OVERWRITE TABLE t2
SELECT `(col_1)?+.+`

输出：

fruits = ["banana", "apple", "mango", "kiwi"]
new_fruits = [fruits[i // 2] if i % 2 == 0 else 'peach' for i in range(len(fruits) * 2)]
print(new_fruits)

如果继续使用['banana', 'peach', 'apple', 'peach', 'mango', 'peach', 'kiwi', 'peach']方法itertools，功能似乎是最快的：

itertools.zip_longest

时间：

new_fruits = [v for v in itertools.zip_longest(fruits, add, fillvalue=add[0])]

Answer 3

列表理解+拼合选项-无需导入：

fruits = ["banana", "apple", "mango", "kiwi"]
add = 'peach'
out = [i for s in [[f, add] for f in fruits] for i in s]
out
# ['banana', 'peach', 'apple', 'peach', 'mango', 'peach', 'kiwi', 'peach']

itertools产品选项：

fruits = ["banana", "apple", "mango", "kiwi"]
add = ["peach"]
out = list(itertools.chain(*itertools.product(fruits, add)))
out
# ['banana', 'peach', 'apple', 'peach', 'mango', 'peach', 'kiwi', 'peach']

一些时间：

%timeit list(chain.from_iterable(zip(fruits, repeat(*add))))
910 ns ± 19.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit list(itertools.chain(*itertools.product(fruits, add)))
1.07 µs ± 13.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit [fruits[i // 2] if i % 2 == 0 else "peach" for i in range(len(fruits) * 2)]
1.62 µs ± 40.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit [i for s in [[f, *add] for f in fruits] for i in s]
2.16 µs ± 630 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

但是，最快的选择似乎是最简单的：

def func(fruits):
    result = []
    for x in fruits:
        result.append(x)
        result.append("peaches")
    return result

%timeit func(fruits)
746 ns ± 15.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Answer 4

与您完全相同，但在第一个列表上：

this.props.convert_point

每隔一个元素向列表中添加一个元素

4 个答案: